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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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234 0 CHAPTER 14 PARTIAL DERIVATIVES<br />

f( -1.951921, - 0.545867, 0.119973) ~ -0.0688, !(0.155142, 0.904622, 0.950293) ~ 0.4084,<br />

!(1.138731, 1.768057, -0.573138) ~ 9.7938. Thus the maximum is approximately 9.7938, and the minimum is<br />

approximately -5.3506.<br />

47. (a) We wish to maximize j(x1, x2, ... , xn) = ytx1x2 · · · Xn subject to<br />

g(x1,x2, .. . , Xn) = X1 +x2 + · · · +xn = c and X.;> 0.<br />

\1 f = ( ~ (X!X2 · .. Xn).; -l(X2 .. · Xn) , ~(X1X2 .. · Xn)'~-l (X!X3 · .. Xn), ... , ~(X t X2 · .. Xn) -!,-l (xl · .. Xn-1) ~<br />

and >. V g = (>., >., ... , >.),so we ne<strong>ed</strong> to solve the system of equations<br />

1/n 1/n 1/n ,<br />

x 1 x 2 · · · Xn = nAx1<br />

1 ( ).l.-1(<br />

;;: XtX2 · • • Xn " XtX3 • · • Xn) = A<br />

1/n 1/n 1/n \<br />

X 1 X 2 · · · Xn = nAX2<br />

This implies n>.x1 = n>.x2 = · · · = n>.x ... Note >. =I 0, otherwise we can't have all x.; > 0. Thus X1 = x2 -= · · · = Xn.<br />

But X ! + X2 + · · · + Xn = c => nx1 = c => X1 = ~ = X2 = X3 = · · · = Xn - Then the only point where f can<br />

n<br />

have an extreme value is (~ , ~, ... , ~). Since we can choose yalues for ( Xt, x2, ... , Xn) that make f as close to<br />

n n n<br />

zero (but not equal) as we like, f has no minimum value. Thus the maximum value is<br />

t(;,;, .... ;) = \};-; ..... ;=;.<br />

(b) From part (a), ~ i s the maximum value of f. Thus j(x1; x2, . .. , x n ) = ytx1x2 · · · Xn :::; ~-<br />

n<br />

,-,-,---,------~· -- X1 + X2 + ··• + Xn . .<br />

x 1 + x2 + ··· + Xn = c, so ytx1X2 · · · Xn :::;<br />

n<br />

n<br />

But<br />

. These two means are equal when f attams 1ts<br />

maximum value ~' but this can occur only at the point(;,*' . .. , * ) we found in part (a). So the means are equal only<br />

- c<br />

when Xt = X2 = X3 = · · · = Xn = -.<br />

- n<br />

14 Review<br />

CONCEPT CHECK<br />

1. (a) A function f of two variables is a rule that assigns to each order<strong>ed</strong> pair (x, y) of real numbers in its domain a unique real<br />

number denot<strong>ed</strong> by f(x, y).<br />

(b) One way to visualize a function of two variables is by graphing it, resulting in the surface z = f(x, y). Another method for<br />

visualizing a function of two variables is a contour map. The contour map consists of level curves ofthe function which are<br />

horizontal traces of the graph of the function project<strong>ed</strong> onto the xy-plane. Also, we can use an arrow diagram such as<br />

Figure I in Section 14.1.<br />

© 2012 Cengage Le'ilming. All Rights Reserv<strong>ed</strong>. May nor be scnnncd. copi<strong>ed</strong>. or duplicntc:d, or postr..'tl ton publicly accessible wch~itc, in whole or in part.

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